Question 28

The factors of $x^{2} - 4$ are
a) $(x - 2), (x - 2)$
b) $(x + 2), (x - 2)$
c) $(x + 2), (x + 2)$
d) $(x - 4), (x - 4)$

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Solution

b) $(x + 2), (x - 2)$

$\to$ We have,
$x^{2} - 4 = x^{2} - 2^{2} = (x + 2) (x - 2)$ $[∵ a^{2} - b^{2} = (a + b) (a - b)]$
Hence, $(x + 2), (x - 2)$ are factors of $x^{2} - 4.$

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